Find the sum $$ 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + n \cdot n! $$
Attempt
$$ 2 \cdot 2! + 3 \cdot 3! = 2 \cdot 2! + 3^{2} \cdot 2! = 2! (2+3^{2})$$
$$ 4 \cdot 4! + 5 \cdot 5! = 4 \cdot 4! + 5^{2} \cdot 4! = 4! (4+5^{2})$$
$$ 6 \cdot 6! + 7 \cdot 7! = 6 \cdot 6! + 7^{2} \cdot 6! = 6! (6+7^{2})$$
then $$ 2 \cdot 2! + 3 \cdot 3! + 4 \cdot 4! + 5 \cdot 5! + 6 \cdot 6! + 7 \cdot 7!$$ $$ = 2! (2+3^{2}) + 4! (4+5^{2}) + 6! (6+7^{2})$$
or perhaps we can calculate in total, for example
$$ 2 \cdot 2! + 3 \cdot 3! + 4 \cdot 4! + 5 \cdot 5!$$ $$ = 2!(2 + 3^{2}) + 4 \cdot 4! + 5 \cdot 5!= 2!(2 + 3^{2}) + 2! (4 \cdot 4 \cdot 3) + 5 \cdot 5! $$ $$ = 2! (2 + 3^{2} + 3 \cdot 4^{2}) + 2! (4 \cdot 3 \cdot 5^{2})$$ $$ = 2! (2 + 3^{2} + 3 \cdot 4^{2} + 3 \cdot 4 \cdot 5^{2})$$
HINT:
$n\cdot n!=(n+1-1)n!=(n+1)!-n!$
Do you know about Telescoping Series?